To find the real zeroes of the polynomial function \( f(x) = x^4 + 6x^3 - 3x^2 - 52x - 60 \), we can use a few methods including the Rational Root Theorem and synthetic division, or numerical methods if necessary.
Let's check for potential rational roots using the Rational Root Theorem. The possible rational roots (factors of the constant term over factors of the leading coefficient) are: \( \pm 1, \pm 2, \pm 3, \pm 4, \pm 5, \pm 6, \pm 10, \pm 12, \pm 15, \pm 20, \pm 30, \pm 60 \).
We can test these values:
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Testing \( x = -5 \): \[ f(-5) = (-5)^4 + 6(-5)^3 - 3(-5)^2 - 52(-5) - 60 \] \[ = 625 - 750 - 75 + 260 - 60 = 0 \] So, \( x = -5 \) is a root.
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Perform synthetic division of \( f(x) \) by \( x + 5 \): \[ \begin{array}{r|rrrrr} -5 & 1 & 6 & -3 & -52 & -60 \ & & -5 & -5 & 40 & 60 \ \hline & 1 & 1 & -8 & -12 & 0 \ \end{array} \] This gives us the quotient \( x^3 + x^2 - 8x - 12 \).
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Now we can find the zeroes of the cubic polynomial \( x^3 + x^2 - 8x - 12 \). We can apply the rational root theorem or just try suspected factors:
- Testing \( x = -3 \): \[ f(-3) = (-3)^3 + (-3)^2 - 8(-3) - 12 = -27 + 9 + 24 - 12 = -6 \quad(\text{not a root}) \]
- Testing \( x = -2 \): \[ f(-2) = (-2)^3 + (-2)^2 - 8(-2) - 12 = -8 + 4 + 16 - 12 = 0 \quad(\text{a root}) \]
Now perform synthetic division with \( x + 2 \): \[ \begin{array}{r|rrrr} -2 & 1 & 1 & -8 & -12 \ & & -2 & 2 & 12 \ \hline & 1 & -1 & -6 & 0 \ \end{array} \] This gives us \( x^2 - x - 6 = 0 \).
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Factor the quadratic: \[ (x - 3)(x + 2) = 0 \]
From this, we can find the roots:
- \( x + 2 = 0 \) gives \( x = -2 \) (already found),
- \( x - 3 = 0 \) gives \( x = 3 \).
The roots of the polynomial are:
- \( x = -5, -2, 3 \).
Now, let's check the options:
- The roots are \( -5, -2, 3 \).
So the correct option is: negative 5, negative 2, 3.